Correlations

$C(\overrightarrow{r},\tau )=⟨\varphi (\overrightarrow{x},t)\varphi (\overrightarrow{x}+\overrightarrow{r},t+\tau )⟨$.

Equal time correlation: $C(\overrightarrow{r},\tau =0)=C(\overrightarrow{r})=⟨\varphi (\overrightarrow{x},t)\varphi (\overrightarrow{x}+\overrightarrow{r},t)⟩$. Note this is averaged over both space and time, but there is no difference in time. $\underset{r\to \mathrm{\infty }}{lim}C(\overrightarrow{r})=C(\mathrm{\infty },0)=⟨\varphi (\overrightarrow{x},t)⟩⟨\varphi (\overrightarrow{x}+\overrightarrow{r},t)⟩$ since they are independent and uncorrelated when far away. For our ising model, $=m\cdot m=m^2$.

Then, the connected correlation function, $C^C(\overrightarrow{r},\tau )=⟨(\varphi (\overrightarrow{x},t)-⟨\varphi ⟩)(\varphi (\overrightarrow{x}+\overrightarrow{r},t+\tau )-⟨\varphi ⟩)⟩$, i.e. this only contains fully connected terms (c.f. Feynman diagram terms in the perturbation theory). In general, these can be measured. Example: X-ray scattering: $I\propto |{\rho }_e(\overrightarrow{k}){|}^2$ so

$$\begin{array}{rl}|\tilde{\rho }(\overrightarrow{k}){|}^2& =\tilde{\rho }(\overrightarrow{k}{)}^{\ast }\tilde{\rho }(\overrightarrow{k})\\ & =\int d{x}^{\prime }\exp (i\overrightarrow{k}\cdot {\overrightarrow{x}}^{\prime })\rho ({\overrightarrow{x}}^{\prime })\int dx\exp (-i\overrightarrow{k}\cdot \overrightarrow{x})\rho (\overrightarrow{x})\\ & =\iint d{x}^{\prime }dx\exp (i\overrightarrow{k}\cdot \overrightarrow{r})\rho ({\overrightarrow{x}}^{\prime })\rho (\overrightarrow{x})\\ & =\int d{r}^{\prime }\exp (i\overrightarrow{k}\cdot \overrightarrow{r})V⟨\rho ({\overrightarrow{x}}^{\prime })\rho ({\overrightarrow{x}}^{\prime }+\overrightarrow{r})⟩\\ & =V\int d{r}^{\prime }\exp (i\overrightarrow{k}\cdot \overrightarrow{r})C(\overrightarrow{r},0)\\ \text{(1)}& & =V\tilde{C}(k).\end{array}$$

White noise implies uncorrelated sources.

$C(r,\tau )$ for an ideal gas. $F^{id}(\rho (\overrightarrow{x}),T)=\rho kT(\ln (\rho {\lambda }^3)-1)$.

So, for small fluctuations, $F\approx F_0+{\mu }_0(\rho -{\rho }_0)+\frac{1}{2}\alpha (\rho -{\rho }_0{)}^2$ Then, integrating these we get $\int Fdx\approx \frac{1}{2}\alpha (\rho -{\rho }_0)+\mathbb{C}$. Thus, for small density fluctuations, $F(\rho )\approx \frac{1}{2}\alpha (\rho -{\rho }_0{)}^2$. Breaking up space to some boxes of $\mathrm{\Delta }V$, $P\sim \exp (-\frac{\frac{1}{2}\alpha (\rho -{\rho }_0{)}^2\mathrm{\Delta }V}{kT})=\exp (-\frac{(\rho -{\rho }_0{)}^2}{\frac{2}{\alpha \beta \mathrm{\Delta }V}})$ Then, ${\sigma }^{-2}=\alpha \beta \mathrm{\Delta }V$.

$⟨(\rho -{\rho }_0{)}^2⟩=\frac{(N-⟨N⟩{)}^2}{(\mathrm{\Delta }V{)}^2}=\frac{N}{(\mathrm{\Delta }V{)}^2}=\frac{{\rho }_0}{\mathrm{\Delta }V}=\frac{1}{\frac{{\rho }_0{P}_0}{{\rho }_0{\rho }_0{P}_0}\mathrm{\Delta }V}=\frac{1}{\frac{P_0}{{\rho }_0{P}_{0}kT}\mathrm{\Delta }V}=\frac{1}{\alpha \beta \mathrm{\Delta }V}$. $C^{id}(\overrightarrow{r},0)=\{\begin{array}{ll}0& r>0\\ \frac{1}{\alpha \beta }& r=0\\ \frac{\rho }{\mathrm{\Delta }V}& \text{otherwise}\end{array}=\frac{1}{\alpha \beta }\delta (\overrightarrow{r})$. Hence, we get white noise, whose fourier transform is a flat distribution.

From last term, ${\sigma }^2\propto \sqrt{N}$ and $\frac{\sigma }{N}\sim \frac{1}{\sqrt{N}}$. Then, $p(m)\propto \exp (-\frac{m^2}{N})$.

Time evolution of densities $\Rightarrow$ diffusion equation. Time evolution of random density fluctuations$({\rho }_i)$ $\to \frac{\partial }{\partial t}{(\rho (\overrightarrow{x},t))}_{{\rho }_i}=D{\mathrm{\nabla }}^2{(\rho (\overrightarrow{x},t))}_{{\rho }_i}$ Onsager Regression Hypothesis.

Want: $C^C(\overrightarrow{r},t)=⟨(\rho (\overrightarrow{x}+\overrightarrow{r},t+\tau )-{\rho }_0)(\rho (\overrightarrow{x},t)-{\rho }_0){⟩}_{ev}$: averages $⟨\cdot ⟩$ ensemble average, no time dependence, $⟨\cdot {⟩}_{eq}$ is equilibrium, $⟨\cdot {⟩}_{ev}$ with time dependence, $[\cdot {]}_i$ noisy average over initial condition.

$C^C(\overrightarrow{r},t)=⟨(\rho (\overrightarrow{x}+\overrightarrow{r},t+\tau )-{\rho }_0)(\rho (\overrightarrow{x},t)-{\rho }_0){⟩}_{ev}$: Use translational invariance in space and time, thus we can set $t$ and $\overrightarrow{x}$ to zero. So, $C^C(\overrightarrow{r},t)=⟨(\rho (\overrightarrow{r},\tau )-{\rho }_0)(\rho (0,0)-{\rho }_0){⟩}_{ev}$

$f[\rho (x)]=f_0+{\frac{\delta f}{\delta \rho }}_{{\rho }_0}(\rho -{\rho }_0)+\frac{1}{2}{\frac{{\delta }^{2}f}{\delta {\rho }^2}}_{{\rho }_0}(\rho -{\rho }_0{)}^2$.

$\frac{\delta f}{\delta \rho }=\frac{\partial f}{\partial \rho }=\mu$.

$\frac{{\delta }^{2}f}{\delta {\rho }^2}=\frac{\delta \mu }{\delta \rho }=\frac{\partial }{\partial \rho }(kT\ln \rho +3kT\ln \lambda )=\frac{kT}{{\rho }_0}=\frac{P}{{\rho }_0^2}$.

$\int f[\rho ]dx=V{\mathcal{F}}_0+\mu \int (\rho -{\rho }_0)dx+\frac{1}{2}\int \alpha (\rho -{\rho }_0{)}^{2}dx\Rightarrow {\mathcal{F}}_{eff}[\rho ]=\frac{1}{2}\alpha (\rho -{\rho }_0{)}^2$.

So, $p(\rho )\propto \exp (-\beta \int \frac{\alpha }{2}(\rho -{\rho }_0{)}^{2}dx)$ Then for a some small volume, $p_j(\rho )\propto \exp (-\beta \alpha (\rho -{\rho }_0{)}^2\mathrm{\Delta }V/2)$. From this, we wanted $C(r,\tau )=⟨(\rho (x+r,t+\tau )-{\rho }_0)(\rho (x,t)-{\rho }_0)⟩$. Since we have a translationally invariant system, and we set time to zero, we get

$$\begin{array}{rl}C(r,0)& =⟨(\rho (r,\tau )-{\rho }_0)(\rho (0)-{\rho }_0){⟩}_{ev}\\ \text{(2)}& & =⟨([\rho (r,\tau ){]}_i-{\rho }_0)({\rho }_i(0)-{\rho }_0){⟩}_{eq}.\end{array}$$

Time evolving,

$$\begin{array}{rl}\frac{\partial C}{\partial t}& =⟨(\frac{\partial }{\partial t}[\rho {]}_i-{\rho }_0)({\rho }_i(0)-{\rho }_0)⟩\\ & =⟨(D{\mathrm{\nabla }}^2[\rho {]}_i-{\rho }_0)({\rho }_i(0)-{\rho }_0)⟩\\ & =D{\mathrm{\nabla }}^2⟨([\rho {]}_i-{\rho }_0)({\rho }_i(0)-{\rho }_0)⟩\\ \text{(3)}& & =D{\mathrm{\nabla }}^{2}C(\bar{r},t).\end{array}$$

$C^{id}(r,\tau =0)=\frac{1}{\beta \alpha }\delta (\overrightarrow{r})$. Then, $\frac{1}{\beta \alpha }G(\overrightarrow{r},\tau )=\frac{1}{\beta \alpha }{\left(\frac{1}{\sqrt{4\pi D\tau }}\right)}^3\exp (-\frac{r^2}{4D\tau })$.